\(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\sqrt{\left(2x\right)^2-2\cdot2x\cdot1+1}=3\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\)
\(=\left|2x-1\right|=3\)
TH1: \(\left|2x-1\right|=2x-1\) với \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)
Pt trở trành:
\(2x-1=3\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\left(tm\right)\)
TH2: \(\left|2x-1\right|=-\left(2x-1\right)\) với \(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)
Pt trở thành:
\(-\left(2x-1\right)=3\) (ĐK: \(x< \dfrac{1}{2}\))
\(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=-1\left(tm\right)\)
Vậy: \(S=\left\{2;-1\right\}\)
Đk: \(x\in R\)
Pt \(\Leftrightarrow4x^2-4x+1=9\)
\(\Leftrightarrow4x^2-4x-8=0\)
\(\Leftrightarrow4x^2-8x+4x-8=0\)
\(\Leftrightarrow4x\left(x-2\right)+4\left(x-2\right)=0\)
\(\Leftrightarrow\left(4x+4\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-1;2\right\}\)
